Q: Crux problem #39 with vectors approach 
It's given a point $P$ inside an equilateral triangle $ABC$ such as segment lengths $PA$, $PB$, $PC$ are $3$, $4$ and $5$ units respectively. Calculate the area of
$\triangle ABC$.


So, let $b:=\overrightarrow{AB},\,c:=\overrightarrow{AC},\,
\overrightarrow{AP}=xb+yc$, 
$$\begin{cases}
(\overrightarrow{AP})^2&=x^2b^2+y^2c^2+2xybc=9\\
(\overrightarrow{AP}-\overrightarrow{AB})^2&=(x-1)^2b^2+y^2c^2+2(x-1)ybc=16\\
(\overrightarrow{AP}-\overrightarrow{AC})^2&=x^2b^2+(y-1)^2c^2+2x(y-1)bc=25
\end{cases}$$
As we have equilateral triangle, $b^2=c^2=2bc=:\frac{1}{a^2}$, then
$$\begin{cases}
x^2+y^2+xy=9a^2\\
(x-1)^2+y^2+(x-1)y=16a^2\\
x^2+(y-1)^2+x(y-1)=25a^2
\end{cases}$$
$$\begin{cases}
x^2+y^2+xy=9a^2\\
-2x+1-y=7a^2\\
-2y+1-x=16a^2
\end{cases}$$
$$\begin{cases}
x^2+y^2+xy=9a^2\\
-2x+1-7a^2=y\\
-2(-2x+1-7a^2)+1-x=16a^2
\end{cases}$$
$$\begin{cases}
9x^2+9y^2+9xy=81a^2\\
-2x+1-7a^2=y\\
3x-1=2a^2
\end{cases}$$
$$\begin{cases}
(2a^2+1)^2+(1-25a^2)^2+(2a^2+1)(1-25a^2)=81a^2\\
y = \frac{1 - 25 a^2}{3}\\
x=\frac{2a^2+1}{3}
\end{cases}$$
$$193 a^4 - 50 a^2 +1=0$$
$$\left[
\begin{array}{l}
\begin{cases}
a^2 = \frac{1}{193}\left(25 - 12 \sqrt{3}\right)\\
x = \frac{1}{193}\left(81 - 8 \sqrt{3}\right)\\
y = \frac{4}{193}\left(25 \sqrt{3} - 36\right)
\end{cases}\\
\begin{cases}
a^2 = \frac{1}{193}\left(25 + 12 \sqrt{3}\right)\\
x = \frac{1}{193}\left(81 + 8 \sqrt{3}\right)\\
y = -\frac{4}{193}\left(25 \sqrt{3} + 36\right)
\end{cases}
\end{array}
\right.$$
The second case doesn't fit as $y<0$, then we have 
$$S_{ABC}=\frac{\sqrt{3}}{4a^2}=\frac{1}{4} \left(36 + 25 \sqrt{3}\right)$$
Although there are two solutions given in [Crux Vol. 1, No. 7][1] (pp. 64-66), I wonder is there more neat geometrical solution?

A: Rotate The whole figure about $B$ anticlockwise by $\pi/3$ radians and assume  $P$ moves to $P'$ then $PP'B$ is equilateral and since $C$ moves to $A$ ,   $APP'$ is a $3-4-5$ triangle.
Now you easily calculate $AB$ from $AP'B$ by cosine rule.

A: Set the triangle with $A(0,0)$, $B(0,z)$ and $ C(z/2, z\sqrt{3}/2)$, where $z$ is the side length (this is the same orientation of the figure in the OP). Call $x$ and $y$ the coordinates of $P$. Now you can solve the problem nearly in one step by solving the following system in three variables, where the equations express the distances of $P$ from the vertices:
$$\displaystyle\left\{ \begin{array}{ll}
                  (x^2+y^2)=9\\
                  (z-x)^2+y^2=16\\
                  (z/2-x)^2+(z\sqrt{3}/2-y)^2=25
                \end{array}
              \right.
 $$
Solving the system and taking the only solution where $x$, $y$, and $z$ are all positive  directly leads, after some simplification of the radicals, to
$$z = (14 \sqrt{3} - 3) \sqrt{\frac{1}{579} (31 + 16 \sqrt{3})}\\
=  \sqrt{25+12\sqrt{3}}$$
as confirmed by WA here and here.
From this we easily obtain
$$S= \frac{\sqrt{3}}{4}  (25+12\sqrt{3})\\
=\frac{1}{4} (36 + 25 \sqrt{3})  $$
